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[parent] proof of dominated convergence theorem (Proof)

It is not difficult to prove that $ f$ is measurable. In fact we can write

$\displaystyle f(x)=\sup_n \inf_{k\ge n} f_k(x) $
and we know that measurable functions are closed under the $ \sup$ and $ \inf$ operation.

Consider the sequence $ g_n(x)=2\Phi(x) - \vert f(x)-f_n(x)\vert$. Clearly $ g_n$ are nonnegative functions since $ f-f_n\le 2\Phi$. So, applying Fatou's Lemma, we obtain

$\displaystyle {\lim_{n\to\infty} \int_X \vert f-f_n\vert\, d\mu \le \limsup_{n\to \infty} \int_X \vert f-f_n\vert\, d\mu}$
  $\displaystyle =$ $\displaystyle - \liminf_{n\to\infty} \int_X -\vert f-f_n\vert\, d\mu$  
  $\displaystyle =$ $\displaystyle \int_X 2\Phi\, d\mu - \liminf_{n\to\infty}\int_X 2\Phi-\vert f-f_n\vert\,d\mu$  
  $\displaystyle \le$ $\displaystyle \int_X 2\Phi\, d\mu - \int_X 2\Phi - \limsup_{n\to \infty}\vert f-f_n\vert\, d\mu$  
  $\displaystyle =$ $\displaystyle \int_X 2\Phi\, d\mu - \int_X 2\Phi\, d\mu = 0.$  



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Cross-references: Fatou's lemma, functions, sequence, operation, closed under, measurable functions, measurable

This is version 1 of proof of dominated convergence theorem, born on 2003-03-07.
Object id is 4077, canonical name is ProofOfDominatedConvergenceTheorem.
Accessed 3746 times total.

Classification:
AMS MSC: 28A20 (Measure and integration :: Classical measure theory :: Measurable and nonmeasurable functions, sequences of measurable functions, modes of convergence)
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